Heisenberg Uncertainty Principle Let’s find an electron Photon changes the momentum of electron x   p  h/ (smaller , bigger p) xp > h/4 x – uncertainty of position p - uncertainty of momentum Et > h/4 E - uncertainty of energy t - uncertainty of time if x = 3.5 ± 0.2 m, x = 0.4 m – it’s the range

Heisenberg Uncertainty Principle xp > h/4 Et > h/4 Strange quantum effects: Observation affects reality Energy is not conserved (for t) Non determinism Quantum randomness Quantum electrodynamics

Example: What is the uncertainty in the position of a 0 Example: What is the uncertainty in the position of a 0.145 kg baseball with a velocity of 37.0 ± 0.3 m/s. (x = 6.0607E-34 m) The uncertainty of momentum is (0.145 kg)(0.6 m/s) = 0.087 kg m/s And now we use xp > h/4: x(0.087 kg m/s) = (6.626E-34 Js)/(4), x = 6.0607E-34 m so x is ±3.03E-34 m So not really very much.

For what period of time is the uncertainty of the energy of an electron 2.5 x 10-19 J? Et > h/4 (2.5 x 10-19 J)t > h/4 t = 2.1 x 10-16 s t = 2.1 x 10-16 s

you know an electron’s position is ± you know an electron’s position is ±.035 nm, what is the minimum uncertainty of its velocity? (4) xp > h/4 p = mv m = 9.11 x 10-31 kg x = 0.070 x 10-9 m (.070 x 10-9 m)p > (6.626 x 10-34 Js)/4 p = 7.5 x 10-25 kg m/s p = mv (7.5 x 10-25 kg m/s) = (9.11 x 10-31 kg)v v = 8.3 x 105 m/s v = 8.3 x 105 m/s